Inverse Discrete-Time Fourier Transform

The inverse discrete-time Fourier transform (IDTFT) is defined as the process of finding the discrete-time sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ from its frequency response X(ω).

Mathematically, the inverse discrete-time Fourier transform is defined as −

$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\: \frac{1}{2\pi}\int_{-\pi}^{\pi}\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\mathit{e}^{\mathit{j\omega n}}\:\mathit{d\omega}\:\:\:\:\:\:...(1)}$$

The solution of the equation (1) for $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ is useful for the analytical purpose, but it is very difficult to evaluate for typical functional forms of function X(ω). Therefore, an alternate method of determining the values of the discrete-time sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ follows directly from the definition of the Fourier transform, i.e.,

$$\mathrm{\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\:\mathrm{=}\:\sum_{n=-\infty }^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{e}^{-\mathit{j\omega n}}\:\mathrm{=}\:...\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{-3}\right)}\mathit{e}^{\mathit{j}\mathrm{3}\omega}\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{-2}\right)}\mathit{e}^{\mathit{j}\mathrm{2}\omega}\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{-1}\right)}\mathit{e}^{\mathit{j}\omega}\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{0}\right)}\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{1}\right)}\mathit{e}^{\mathit{-j}\omega}\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{2}\right)}\mathit{e}^{\mathit{-j}2\omega}\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{3}\right)}\mathit{e}^{\mathit{-j}3\omega}\:\:\:\:\:\:...(2)}$$

Hence, from the equation of X(ω) we can say that, if X(ω) can be expressed as a series of complex exponentials as given in the equation (2), then $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ is simply the coefficient of $\mathit{e}^{-\mathit{j\omega n}}$.

Numerical Example (1)

Find the inverse discrete-time Fourier transform of X(ω).

$$\mathrm{\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\:\mathrm{=}\:\mathit{e}^{-\mathit{j}\omega}\:;\mathrm{for}\:-\pi \leq \omega \leq \pi}$$

Solution

The given Fourier transform is,

$$\mathrm{\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\:\mathrm{=}\:\mathit{e}^{-\mathit{j}\omega}}$$

From the definition of inverse discrete-time Fourier transform, we have,

$$\mathrm{\mathit{F}^{-1}\mathrm{\left[ \mathit{X}\mathrm{\left(\mathit{\omega}\right)}\right ]}\:\mathrm{=}\:\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\frac{1}{2\pi}\int_{-\pi }^{\pi }\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\mathit{e}^{\mathit{j\omega n}}\:\mathit{d\omega}}$$

$$\mathrm{\therefore \mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\frac{1}{2\pi}\int_{-\pi}^{\pi}\mathit{e}^{-\mathit{j}\omega }\mathit{e}^{\mathit{j}\omega n}\:\mathit{d\omega}\:\mathrm{=}\:\:\frac{1}{2\pi}\int_{-\pi}^{\pi}\mathit{e}^{\mathit{j}\omega\mathrm{\left( \mathit{n}-1\right )} }\:\mathit{d\omega}}$$

$$\mathrm{\Rightarrow \mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\frac{1}{2\pi}\mathrm{\left[ \frac{\mathit{e}^{\mathit{j}\omega\mathrm{\left( \mathit{n}-1\right)}}}{\mathit{j}\mathrm{\left( \mathit{n}-1\right)}} \right]}^{\pi}_{-\pi}\:\mathrm{=}\:\frac{1}{2\pi}\mathrm{\left[\frac{\mathit{e} ^{\mathit{j\pi}\mathrm{\left(\mathit{n}-1\right)}}-\mathit{e}^{-\mathit{j\pi \mathrm{\left ( n-1 \right )}}}}{\mathit{j}\mathrm{\left ( \mathit{n}-1 \right )}}\right ]}}$$

$$\mathrm{\Rightarrow \mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\frac{1}{\pi \mathrm{\left(\mathit{n}-1 \right)}}\mathrm{\left[\frac{\mathit{e} ^{\mathit{j\pi}\mathrm{\left(\mathit{n}-1\right)}}-\mathit{e}^{-\mathit{j\pi \mathrm{\left ( \mathit{n}-1 \right )}}}}{2\mathit{j}} \right ]}\:\mathrm{=}\:\frac{sin\:\pi \mathrm{\left(\mathit{n}-1 \right )}}{\pi\mathrm{\left( \mathit{n}-1 \right )}}}$$

$$\mathrm{\therefore \mathit{F}^{-1}\mathrm{\left[ \mathit{X}\mathrm{\left(\mathit{\omega}\right)}\right ]}\:\mathrm{=}\:\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\frac{sin\:\pi \mathrm{\left(\mathit{n}-1 \right )}}{\pi\mathrm{\left( \mathit{n}-1 \right )}}}$$

Numerical Example (2)

Determine the function $\mathit{x}\mathrm{\left(\mathit{n}\right)}$, i.e., inverse discrete-time Fourier transform of X(ω).

$$\mathrm{\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\:\mathrm{=}\:4\:\mathrm{+}\:\mathit{e}^{-\mathit{j}\omega}\:\mathrm{+}\:2\mathit{e}^{-\mathit{j}2\omega}\:\mathrm{+}\:4\mathit{e}^{-\mathit{j}4\omega}}$$

Solution

The given Fourier transform is,

$$\mathrm{\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\:\mathrm{=}\:4\:\mathrm{+}\:\mathit{e}^{-\mathit{j}\omega}\:\mathrm{+}\:2\mathit{e}^{-\mathit{j}2\omega}\:\mathrm{+}\:3\mathit{e}^{-\mathit{j}3\omega}\:\:\:\:\:\:...(3)}$$

Now, from the definition of the discrete-time Fourier transform, we have,

$$\mathrm{\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\:\mathrm{=}\:\sum_{n=-\infty }^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{e}^{-\mathit{j}\omega n}\:\mathrm{=}\:...\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{-3}\right)}\mathit{e}^{\mathit{j}\mathrm{3}\omega}\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{-2}\right)}\mathit{e}^{\mathit{j}\mathrm{2}\omega}\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{-1}\right)}\mathit{e}^{\mathit{j}\omega}\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{0}\right)}\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{1}\right)}\mathit{e}^{-\mathit{j}\omega }\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{2}\right)}\mathit{e}^{-\mathit{j}2\omega }\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{3}\right)}\mathit{e}^{-\mathit{j}3\omega }\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{4}\right)}\mathit{e}^{-\mathit{j}4\omega }\:\:\:\:\:\:...(4)}$$

Comparing equations (3) and (4), we get,

$$\mathrm{\mathit{x}\mathrm{\left(\mathrm{0}\right)}\:\mathrm{=}\:4;\mathit{x}\mathrm{\left(\mathrm{1}\right)}\:\mathrm{=}\:1;\mathit{x}\mathrm{\left(\mathrm{2}\right)}\:\mathrm{=}\:2;\mathit{x}\mathrm{\left(\mathrm{3}\right)}\:\mathrm{=}\:0;\mathit{x}\mathrm{\left(\mathrm{4}\right)}\:\mathrm{=}\:4}$$

$$\mathrm{\therefore \mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathrm{\left\{4,1,2,0,4 \right\}}}$$